STRUCTURE OF Ba-132, Ba-134, Ba-135, Ba-136, Ba-137, Ba-138
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Nuclear structure of Barium with 30 blank positions Naturally occurring barium (Ba) is a mix of six stable isotopes and one very long-lived radioactive primordial isotope, barium-130, recently identified as being unstable by geochemical means. Especially barium has 4 stable isotopes like Ba-132, Ba-134, Ba-136 and Ba-138 with S =0 and 2 stable isotopes like Ba-135 and Ba-137 with S =+3/2. ' '''Comparing the Barium of 56 protons with Xenon of 54 protons (even number ) we conclude that the structure of Barium has the same high symmetry as that of Xenon. ( See my STRUCTURE OF Xe-124..Xe-134 ). In the following diagram of Ba the additional p55n55 and p56n56 are not shown . However the position of n55 of the vertical n55p55 is found by using the top view of the third horizontal plane. It fills the black positions of the extra 1(n) formed by p41 and p49 (See also the top view of the third h. plane of Xe). Because of symmetry also the blank position of 1(p) is filled by the p56 because the vertical p56n56 fills the symmetrical blank positions of the p55n55. It is of interest to note that the protons p53 and p56 are able to form a new blank position in order to receive 1(n) Under this condition the 8(n) of Tellurium are reduced to 6(n). See my STRUCTURE OF Te-120...Te-130 ) So using all the cases of the top view the extra blank positions able to be filled by extra neutrons of Ba are given by The two squares give 8n The first and sixth plane give 4(n) The second and fifth plane give 4{n} +8n The third and the fourth plane giving the rest 6(n) That is N = 8n + 4(n) + 4{n} + 8n + 6(n) = 30 extra neutrons '''STRUCTURE OF Ba-132, Ba-134, Ba-136, Ba-138 WITH S = 0' Since the p = 56 and n =56 give S = 0 we conclude that the S =0 of Ba is due to the extra neutrons of opposite spins. For example the Ba-132 of 20 extra neutrons has 4{n} + 16n of opposite spins , while the Ba-138 of 26 extra neutrons has 4{n} + 16n + 6(n) of opposite spins. STRUCTURE OF Ba-135 AND Ba-137 WITH S = +2/3 Similarly since the p = p6 and n=56 give S = 0 we conclude that the total S = +3/2 of the Ba-135 of 23 extra neutrons is due to the total S=+3/2 . So Ba-135 has 13 extra neutrons of positive spins and 10 extra neutrons of negative spins. In the same way the Ba-137 of 25 extra neutrons has 14 extra neutrons of positive spins and 11 extra neutrons of negative spins. ' ' DIAGRAM OF Ba FORMING 30 BLANK POSITIONS Here the vertical n55p55 and p56n56 are not shown, but you can see the positions of n55 and p56 by using the top view of the third horizontal plane. Here you see the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins . But you cannot see the additional p49n49 the n52p52 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. In the same way the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Moreover the extra neutrons 4(n) of the first and the sixth plane are not shown, while the extra neutrons 4n existing over the p31 and p32, and under the p21 and p22 with the extra 4n existing near the p23, p24, p29 and p30, are shown. ' n40.......p40.......n ' ' n........p38..........n38 H. Square with n' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth H. plane' ' n........ p29.........n10.........p10…… n30' ' n29………..p9..........n9 …….p30..........n Fifth H. plane' ' p47.......n27.........p8..........n8.........p28......... n48' ' n45.......p27.........n7..........p7........n28..........p46 Fourth H. plane' ' n47......p25.........n6.........p6..........n26...........p48' ' p45......n25……….p5..........n5……….p26.........n46 Third H. plane' ' n23………p4........n4………….p24...........n' ' n........p23……..n3………p3………..n24 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' n.......p37......n37 ' ' n39.....p39..........n ' H. Square with n TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' ' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE Here the n near the p14 fills the blank position formed by p51 and p14. While the {n} near the p14 fills the blank position formed by p14, p24 and p44. The 2n near p24 and p23 fill the blank positions formed by p24 and p48 as well as by p23 and p45. Moreover the blank position of {n} near p23 is formed by p23, p13 and p41. Finally the blank position of n near the p13 is formed byp13 and p49. That is we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' n' ' n14.......p14........{n}' ' n23.......p4........n4.........p24......n' n.......p23........n3........p3.........n24 ' {n}.......p13......n13' ' n ' ' ' ' ' ' ' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' '''HERE YOU SEE THE p53 AND n54 WHICH REPLACE THE 1(p) AND 1(n) . SO THE NUMBER (n) OF THE THIRD AND THE FOURTH PLANE IS 6(n). ALSO THE n55 FILLS THE POSITION OF 1(n) AND THE p37 FILLS THE POSITION OF 1(p). HERE THE p41, n42, n43 AND p44 MAKE THE SQUARE OF THE CENTRAL PARALLELEPIPED, WHILE THE p45 AND n47 ALONG WITH THE n46 AND p48 ARE THE DEUTERONS OF THE TWO ALPHA PARTICLES. AT THE SAME PLANE ARE SHOWN ALSO THE DEUTERONS LIKE n16p16 AND p15n15 ALONG WITH THE ADDITIONAL p49, n52, n50, AND p51 WHICH ARE THE DEUTERONS OF THE TWO ALPHA PARTICLES EXISTING IN FRONT OF THE CENTRAL PARALLELEPIPED AND BEHIND IT. ' (n)........p37.......n50.......p51......(n) ' ' p53........n42........p16......n16......p44.........n54''' ' n47........p25........n6........p6........n26.........p48' ' p45........n25........p5........n5........p26........ n46' ' n55.........p41.......n15.......p15.......n43.........(p)' ' (n)........p49.......n52.......(p)' ' ' Category:Fundamental physics concepts